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\section*{Question 1}
$Prove: \mathbb{P} (A \cup B)=\mathbb{P} (A)+\mathbb{P} (B)-\mathbb{P}(A \cap B) \\
\mathbb{P} (A) = \mathbb{P} (AB^c) + \mathbb{P} (AB) \\
\mathbb{P} (B) = \mathbb{P} (A^cB) + \mathbb{P} (AB) \\
\mathbb{P} (A) + \mathbb{P} (B) = \mathbb{P} (AB^c) + \mathbb{P} (AB) + \mathbb{P} (A^cB) + \mathbb{P} (AB) \\
\mathbb{P} (A \cup B) = \mathbb{P} (AB^c \cup A^cB \cup AB) \\
\mathbb{P} (A \cup B) = \mathbb{P} (AB^c) + \mathbb{P} (A^cB) + \mathbb{P} (AB) \\
Therefore \mathbb{P} (A \cup B) = \mathbb{P} (A) + \mathbb{P} (B) - \mathbb{P} (A \cap B) $

\section*{Question 2}
$ Define: \\
Sent \, 1 = A, Sent \, 0 = B, Recieved \, 1 = C, Recieved \, 0 = D \\
\mathbb{P} (A) = 0.5\\
\mathbb{P} (B) = 0.5\\
\mathbb{P} (C|A) = 0.99\\
\mathbb{P} (D|A) = 0.01\\
\mathbb{P} (C|B) = 0.05\\
\mathbb{P} (D|B) = 0.95\\
\mathbb{P} (A|D) = \dfrac{ \mathbb{P}(A) \times \mathbb{P} (D|A) }{ \mathbb{P} (D)}\\
\mathbb{P} (D) = \mathbb{P}(B \cap D) + \mathbb{P}(A \cap D)\\
\mathbb{P} (D|A) = \dfrac{\mathbb{P}(A \cap D)}{\mathbb{P}(A)}\\
\mathbb{P} (A \cap D) = \mathbb{P} (D|A) \times \mathbb{P}(A) = 0.005\\
\mathbb{P} (D|B) = \dfrac{\mathbb{P}(B \cap D)}{\mathbb{P}(B)}\\
\mathbb{P} (B \cap D) = \mathbb{P} (D|B) \times \mathbb{P}(B) = 0.475\\
\mathbb{P} (D) = 0.48\\
Therefore: \\
\mathbb{P} (A|D) = \dfrac{0.5 \times 0.01}{0.48} = 0.010416666$
\section*{Question 3}
$ \int_{0.25}^{0.75} 2x \, \mathrm{d} x = 0.75^2 - 0.25^2 = 0.5$
\section*{Question 4}
\section*{Question 5}
\section*{Question 6}
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